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Laws of Motion
Solutions to Subjective AssignmentsLEVEL – I
1. We find that the each of the blocks A & B experience following forces.
Forces acting on A :(a) Earths gra!itational field force " weight of the block A " #1g
(b) $or%al contact force exerted b the hori'ontal srface " $1.
(c) ontact force exerted b the block B " $
(d) *he external hori'ontal force " F.
Forces acting on B :
(a) Earths gra!itational field force " weight of the block B " #+g
(b) $or%al contact force exerted b the hori'ontal srface " $+.
(c) ontact force exerted b the block A " $
Both the blocks %o!e together. ,ence both will ha!e sa%e acceleration- let s
chose /axis shown. *here is no 0/axial acceleration as the blocks are not co%ing
off the srface. *he F.B.. diagra% of 2A3 as a sste% is shown in figre (i) . *he
F.B.. of 2B3 as a sste% shown in figre (ii).$1
F$
#1g
a
$+
$
#+g
a
Fig. (i) Fig. (ii)
Fro% figre (i)- $1 4 #1g " 5 6(1)
For A- F 4 $ " #1a 6(+)
Fro% figre (ii) $+ 4 #+g " 5 6(7)
For B- $ " #+a 6(8)
*hs sol!ing (1)- (+)- (7) and (8) We obtain
a "+1
+
+1 ##
F#$and
##
F
+=
+
+. Along !ertical direction-
* cos θ " %g 6(1) Along hori'ontal direction
* sinθ " %a 6(+)9sing (1) and (+) tan θ " ag
%g
* sin θ
* cos θ
a
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∴ a " g tan θ " ;.< tan 75o
"+s:%
7
%s+.
7. *he F.B.. of %1 gi!es-
$1 " %1g os θ 6(1)* " %1g ?in θ 6(+)*he F.B.. of %+- gi!es-
$+ " %+g is α 6(7)* " %+g ?in α 6(8)
*
%1 g sin θ %1 g cos θ
$1
$+
%+ g cos α %+ g sin α
*
%1 %+
*aking the ratio of (+) and (8)-
θα
=?in
?in
%
%
+
1" o
o
@5?in
75?in "
7
1
7
+
+
1=× .
∴7
1
%
%
+
1 = .
8. et the block be displaced to the left b x
the forces acting on the block will be kx to the right b both the springs
ths F " %a or a "%
kx+
%
F=
=. $
8/18/2019 Hints and Solution LOM
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" µk%g " 5.1= × 1 × ;.<" 1.8> $.
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@. F " (%1 D %+)a
∴ a "+=
1
%%
F
+1 +=
+ " >1
%s+
if the string breaks- %+ will %o!e with constant !elocit attained b it while %1 will%o!e with a acceleration
a′ "=1 %s+
>. (a) Angle of repose θ " tan−1(µ)∴ θ " tan−1(5.7) " 1@.>°
(b) Frictional force " µs%gcos(θ+)
∴ f fr " 5.7 × cos
+
1@.>%g " 5.18= %g
8/18/2019 Hints and Solution LOM
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15. *aking the differential ele%ent at an anglar
position θ- we obtain*he net tangential force " ∫ (d%)g sin θ
" θ
∫ α
sing)ds.(l
#
5
θ
θ∫ α
sing)Ed.(l
#
5
" ∫ α
θθ5
dsinl
#Eg
" [ ]αθ− 5cosl
#Eg-
where α " l ⇒ . l*R +⇒ *angential acceleration "
)
E
cos1(Eg
− .
θ dθR
(dm)g
dm
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LEVEL – II
1. *he forces acting on the particle are%g (!erticall downward) %a5 (psedoforce !erticall downward) $ nor%al
reaction- nor%al to the inclined plane)esol!ing the forces along theincline- we ha!e
F " % (a5 D g) sin θ⇒ acceleration down the plane
a " (a5 D g) sin θ "
+ g
+
gsin 755 "
8
g7
istance tra!ersed along the incline
sec θ " +at+
1 "
+t<
g7
t " g77
A1@
g7
75secA<
=
+. #otion of the bod p the plane
F " %g sin θ D µ %g cos θ " %g sin θ D µcos θG#otion of the bod down the plane
F" µ%g cos θ / %g sin θ " %g H( µcos θ / sin θ) HIt µ " 5.+- θ " 755- & % " + kg
mg sinθ
µmg cos θ
F'
mg sinθ
µmg cos θ
m
m
F
'ote / Jf F is negati!e alter its ass%ed direction.
7. Acceleration of the %an relati!e to the ele!ator " + %s+ (pwards) Acceleration of the ele!ator relati!e to earth " 7 %s+ (downwards),ence- acceleration of the %ass relati!e to earth " 7 4 + " 1 %s+ (downwards)onsider free bod diagra% of %an :Jf * " tension in the spring K%g 4 * " %a
⇒ =5 × 15 4 * " =5 × 1⇒ * " 8=5 $.
,ence- extension in the spring " +555
8=5
" 5.++= %.
8. %!ds
d! " αcos
7
%gds "
a
dα
∫ !
5
!d! " αα∫ α
dcosa7
g
5
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α= sina7
g
+
!+
! " αsina7
g+.
=. Acceleration of a block sliding down an inclinedplane is gi!en b
a " g ( / µ cos θ)for + kg block µ1 µ+ - hence a1 C a+,ence the blocks %o!e separatel and with neLalaccelerations. Acceleration of + kg block isa " g(sin 75 4 5.+ cos 75)
" 15 (5.= 4 5.1>7+) " 7.+> %s+
Jn case (b)- when µ1 " 5.7 and µ+ " 5.+- the + kg block has less acceleration thanthe 7 kg block.
*hs the blocks %o!e together. When we draw the FB and resol!ed the forces
along the plane- we get
F " (%1 D %+) g sin θ / (%1µ D %+µ+) g cos θ
⇒ a "+1 %%
F
+ " g sin θ /θ
+µ+µ
cosg%%
)%%(
+1
++11
" 15 sin 75 /@
)+.587.5+( ×+× × 15 × cos 75° " +.;> %s+
@. Jnitiall
%g 4 F " %a ↓ . . . (i) After releasing a %ass (∆%)F 4 (%/ ∆%)g " (% / ∆%) a ↑ . . .(ii)Where F " pward thrst of air
B sol!ing (i) and (ii) & ptting %g " w-
obtain the answer.
a
F
mg
a
F
(m-∆m)g
>. et there is no relati!e %otion between
the blocks B and ,ence * " (%B D %)a (1)
And % Ag / * " % Aa (+)
Fro% (1) and (+)- we get
a "+
B A
A s:%7
=
1<
75
%%%
g%==
++⇒ $et force on the block is
F " %a " 15 × (=7) $ " 1@.@ $
mAg
T
F.B.D. of the blocks
NB+mC
f r
f r
NC
mB g
T
NB
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?ince force acting on the block is onl frictional force.
Jf %axi%% !ale of frictional force acting on is f %ax - then
f (%ax) " µ%Bg " 5.8 × = × 15 " +5 $ F ≤ f %ax ,ence there is no relati!e %otion between the block B and . *herefore
distance %o!ed b is +% onl.
8/18/2019 Hints and Solution LOM
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(c) When F " 185 $
* " >5 $
T
m1
m1g
1
⇒ * / %1g " %1a1 (1)⇒ >5 $ / +5 $ " + × a1 ⇒ a1 " += %s+
yP
y1
m1
y2
m2
F
* / %+g " %+a+ (1)
⇒ >5$ / =5 $ " =a+ ⇒ a+ " 8 %s+ onstraint eLation
I / + D I / 1 " c⇒ +I / 1 / + " c
⇒ 5dt
&d
dt
&d
dt
&d+
+
+
+
+
1
+
I
+
=−−
⇒ aI "+1s:%
+
+;
+
aa+=
+
;. ELation of %otion for %:
ΣFx " %a ⇒ $ " %a 6(1)
ΣF " 5 ?ince % is at rest relation to # at the !erge ofslipping.⇒ f 4 %g " 5
a
$
µ$
%g
ELation of %otion for #:
F 4 µ$′ 4 $ " #a 6(8)
Where $ " µ%g
- a "µ
=µ
ga-
%g and since the block
# does not ha!e !erticall %otion.
⇒ $′ " #g D f Itting !ales of $′- $ and a in (8)- we obtain
F 4 µ (#g D f) 4 $ " #a F / µ (#g D %g) 4 %a " #a⇒ F / µg (# D %) " (# D %)a
$a
F
µ$′
µ$ #g
$′
⇒ a "%#
)%#(gF
++µ−
f ≤ f %ax ⇒ %g ≤ µ$⇒ %g ≤ µ%a
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⇒ g ≤ µ.%#
)%#(gF
++µ−
⇒ (# D %) g ≤ µF / µ+ g (# D %)⇒ µ+g(# D %) / µF D (# D %) g ≤ 5⇒ +µ+ / =µ D + ≤ 5
For critical-+µ+ / =µ D + " 5(µ / +) (+µ / 1) " 5
∴ µ " + or µ "+
1
Iracticall µ does not beco%e +.∴ µ " 5.=.
15. et x1 → distance of the plle (wedge)fro% the wall.
x+ → distance of the bar fro% the plle.*herefore
x1 D x+ " l
+
1+
dt
xd → acceleration of wedge
+
1+
dt
xd "
+
++
dt
xd−
α
ax+
x1ax
&
++
+
dt
xd → acceleration of the bar- w.r.t.
wedge
+
1+
dt
xd " +
++
dt
xd " a (sa).
Acceleration of the bar
a
a cos α
a sin αa
α
$et acceleration of wedge along the
hori'ontal srface " a 6(1)
$et acceleration of bar along the hori'ontal
srface figre (i)
" a 4 co%p. of acceleration of bar w.r.t.
wedge
" a 4 a cos α 6(+)Merticall downward " a sin α 6(7)
$1
*
a
*
$
#g
α
Forces on wedge and bar are shown
separatel in figre (ii) and fig. (iii)
respecti!el (i!) and (!).
Fro% the F.B.. of the wedge
Along the chosen /direction
* D $1 sin α 4 * cos α " #a 6(8)
α
$1
%g
α
a/acos α
a sin α
along 0/direction
For bar
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$1 cos α D * sin α D #g 4 $ " 5 6(=)Fro% the F.B.. of the bar along /direction
cos α 4 $1 sin α " % (a 4 a cos α) 6(@)For Wedge: along 0/direction
%g 4 $1 cos α 4 * sin α " % a sin α 6(>)
$1 cos α * sin α
$1 sin α* cos α
%g
a/a cos α
a sin α
Fro% eLations (8)- (=)- (@) and (>)
eli%inating other Lantities we can find a.*he
proble% can be sol!ed b choosing /0 axis
for the sste% along and nor%al to the
srface of wedge. Nst do b orself.
$
* cos α
#g
$1 cos α
* sin α
$1 sin α*
a
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Solutions to Objective Assignments
LEVEL – I
"+ *+ " %a " %7
F
%7
F=
F 4 *1 " %a
*1 " F / %a
" F /7
F+
7
F=
⇒ .+
1
7:F+
7:F
*
*
1
+ ==
m
a a aZYX
Fm mT1
T1
T2
T2
#+ *he contact force F accelerates the bod
of %ass %1 " + kg with an acceleration
a " s:%1+1
7
%%
F
+1
=+
=+
⇒ F " %1a " 1 × 1 " 1'+
1 kg2 kg
m1F '
m1
m2
F
+ a "%
Fi%pact "%
)dt:d%(! "
+
)1)(=( " "+m*s"+
+ Jn nifor% circlar %otion !F⊥ where F is the centripetal force
⇒ OE " ttancons%!+
1 +
=%+ *he net force F " #1g sinα / #+ g sin β
" (#1 sin α / #+ sin β)gWhen #1 " #+ and α1 " α+ -the net force is 'ero.
⇒ *he acceleration of the sste% " 5
7+ 9sing the expression obtained in Lestion no. 1
We obtain µ " 1/ (1n+)- ptting n " +we obtain µ " &+7
)+ *he acceleration of % along the plane isg sin θ becase the inclined plane is s%ooth⇒ *he co%ponent of the tension * parallelto the inclined plane is 'ero. *he onl
dri!ing force is co%ponent of gra!itational
force %g parallel to the plane- that is eLal
to %g sin θ.*herefore- for the eLilibri% of the bod-
mg cos θ . 0+
θT
m
mg
mg sin θθ mg cosθ
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9+ ?ince block has been pshed and is broght to %otion hence kinetic friction will
act
∴ %gsinθ − f r " %a∴ f r " µ+%gcosθ∴ gsinθ − µ+g cosθ " a
⇒ a " g(sinθ − µ+cosθ)
1&+ a "%ass
Fi%pact "[ ]
%
)dt:d%(!
"%
)!(! α
"%
!+α (n%ericall)
11+ Following the pre!ios
procedreF 4 *+ " %7a
⇒ F 4 *+ " %7 7+1 %%%
F
++
⇒ *+ "7+1
+1
%%%
F)%%(
++
+
⇒ *+ " $;7@)+>
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1+ For the eLilibri% of the bod %.
* cosθ " %gFor hori'ontal acceleration of %:
* sin θ " %a
⇒
%g
%a
cos*
sin*=
θ
θ
⇒ θ " t$n 1 ($*g+
T cos θ
T sin θ
T
θθ
mg
a
1%+ For eLilibri% of the bod (block)
f 4 %g " 5
⇒ f " %g " (5.1) (;.
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⇒ µcot θ " 1 /+n
1
⇒ µ " t$nn
11
"
− .
LEVEL – II
1. Acceleration of #-F
aM
= ÷
+1 F .t+ #
=l
+#t
F=
l.
+. #g 4 * " #a 6(i)
* " %a6(ii)?ol!ing (i) and (ii)
Mga
(M m)=
+ N
a
M g
FB of %an#g 4 $ " #a
#%g$(# %)
= +7. #g * #a− = 6(i)
* " #a 6(ii)
#g a(# #)= +# g
a(# #)
=+
ma sin mg cosθ = θa g cot= θ
T
R
M g
M
#ggcot
(# #)θ =
+cot # cot # #θ + θ =
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#cot#
(1 cot )
θ=
− θ* " # a
" #. g cot θ#g
* tan= θ .m a s i n θ
m a
m g s i n θm g c o s θ
θ
+ m a c o s θ
8. Jf the tendenc of relati!e %otion along the co%%on tangent does not exist- thenco%ponent of contact force along co%%on tangent will be 'ero.
=. For the two !ales of F- i.e.- F " 1=5 $ and F " 1+5 $K the tensions in the string are
F*
+= " >= $ and @5 $. Jn the first case- the accelerations of the two %asses are eLal
and opposite- while in the second the accelerations are 'ero.
@. *1 " +µ %g 6(i)F " ; µ%g 6(ii)
∴ (a)- (b).
2 m g
T2 m gµ
F2 m gµ
2 m gµ5 µ m g
>. kx " 1+5 $-
22 12a ! m " s2
−= .
1 2
2 2 k g
;. ?ince there is relati!e acceleration between the fra%es.?1 %a be at rest ?+ is accelerating
?1 %a be acceleration and ?+ is at rest?1 and ?+ both are accelerating with difference acceleration?1 and ?+ both cannot be at rest.
15. 1 1$ % H a H sin %gcos+ θ = θ
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m g c o s θθ
N + m # a # s i n1 1 θ
m g s i n + m # a # c o sθ θ1
m # a #1
# a #1
[ ]1 1 N m g cos # a # sin= θ− θ1 1 N sin M # a #θ =
FB of %
m g c o s θ
N + m # a # s i n1 1 θ
m g s i n θ
2 1 N mg N cos= + θ
2 1 N mg m(g cos # a # sin )cos= + θ− θ θ
2 N (M m)g∴ < + .
m g
N s i n1 θ
N2
N c o s1 θ
COMPREHENSION
1.+
?
%!%g
r ≤ µ - since friction is static.
+. $or%al reaction decreases- and so the li%iting friction is lower.7. $et force " f 4 " 5.8. to @.
F.B.. of %an
a
N
T
M g
$ D * 4 #g " #aWith $ " #g
⇒ * " #a 6(i)
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F.B.. of box :
a
N
T
( m + m ) g’
T $ N $ mg $ m′g" (% D %′)a 6(ii)Where # " @5 kg-
% " +5 kg- %′ " 75 kg?ol!ing eLation (i) and (ii)- we get
a " 115 %s 41 * " @@55 $.
MA02! 0!E 3OLLO4I'5
1. When bod is at rest then friction is static and if bod is %o!ing then friction on it iskinetic.
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